CTET Maths MCQ - Time and Work

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I. Ratio of speed = 1 : 2 : 4
If equal time is spent on each of these, then ratio of distance = 1 : 2 : 4
So, Sachin runs twice the distance that he jogs. Thus, I is true.
II. He runs at 8 km/hr and therefore walks at 2 km/hr. Hence II is not true.
III. Ratio of speed = 1 : 2 : 4
Ratio of time = 2: 2 : 1
Ratio of distance = 2 x l: 2 x 2 : 4 x 1
= 2: 4: 4
If Sachin covers a total distance of 10 km, then he

Thus, III is true.
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In 1 minute, monkey ascends 10 metres but he takes 1 minute to slip down 2 metres.
Thus, atthe end of 2 minutes, net ascending of the monkey is = 10 - 2 = 8 metres.
Thus, to have a net ascending of 8 metres, process of ascending and then slipping happens once.
So,to cover 64 metres, above process is repeated 64/8 or 8 times. It is clear that in 8 such happenings, the monkey will slip 7 times, because 8th time, he will ascend to the top.
Thus, in climbing 7 times and slipping 7 times, he covers (7 x 8) or 56 metres.
Time taken to cover 56 metres = (56 x 2)/8 = 14 minutes
Remaining distance = 64 - 56 = 8 metres
Time taken to ascend 8 metres = 8/10 = 4/5 min
Total time taken = 14 minutes + 4/5 min
= 14 mins. 48 sec.
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If they meet t hrs after 7 a.m,

36t + 48t - 16 = 68
t = 1 hr.
⇒ They meet at a distance of
⇒ 10 x 60 x 60 = 36000 m i.e., 36 km from A.
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Two equations are

where T1 and T2 are time taken by trains A and B to cover the whole distance
and
Solving equations (i) and (ii), we get
T1 = 10 hrs. and T2 = 9 hrs.
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Refer to the data below and answer the questions that follow.
The variation in the speed of a car on a particular day at the respective times is shown in the table below:

 s(km/hr)04050851010
 t(hr)11.00 am11.30 am1.00 pm1.30 pm3.30 pm4.30 pm

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Graph of speed vs time is plotted as shown. Area under the graph and time axis gives the distance.

Required distance
= A(ΔOAP) + A(ΔBXA) + A(▢APQX)

= 7.5 km.
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